text:
\section{Introduction}
Recently, Lu and Do \cite{DL08} extended the sampling problem assuming
that the signals to be sampled belong to a union of subspaces instead of a single subspace.
This approach represents a new paradigm in sampling theory.

In the classical setting the  signals are assumed to belong to a  single space of functions,
usually the Paley-Wiener space of band-limited functions. Since in many applications
the band-limitedness hypothesis is not realistic, other spaces of functions were considered,
mainly, shift-invariant spaces (SIS) with very ge\-neral generators.
In the approach of Lu and Do, the signals belong to a union of subspaces instead of a single one.
This simple idea  may have a great impact in many applications in signal processing,
in particular in the emerging theory of compressed sensing \cite{CT06}, \cite{CRT06}, \cite{Don06} and signals with finite rate of innovations \cite{VMB02}.

To describe the problem,  assume that ${\mathcal{F}}$ is a union of  subspaces from some Hilbert space
 $\mathcal{H}$ and a signal $s$ is extracted from ${\mathcal{F}}$. We take some measurements of that signal. These measurements can be thought of as the result of the  application of a series of functionals $\{\varphi_{\alpha}\}_{\alpha}$ to our signal $s.$
The problem is then to reconstruct the signal  using only the measurements $\{\varphi_{\alpha}(s)\}_{\alpha}$ and some des\-cription of the subspaces in ${\mathcal{F}}$.
The series of functionals define  an operator, {\it the sampling operator}, acting on the
ambient space $\mathcal{H}$ and taking values in a suitable sequence space.
Under some hypothesis on the structure of the subspaces, Lu and Do found necessary and sufficient conditions on these functionals in order for the sampling operator to be  stable and one-to-one when restricted to the union of the subspaces.
These conditions were obtained in two settings. In the euclidian space and in $L^2({\mathbb{R}}^d)$.
In this latter case the subspaces considered were finitely generated shift-invariant spaces.

Blumensath and Davies \cite{BD07} studied the problem of sampling in union of subspaces  in the finite dimensional case, extending some of  the results in  Lu and Do \cite{DL08}.  They applied their results to  compressed sensing models and sparse signals.
In \cite{EM08}, Eldar developed a general framework for robust and efficient recovery of a signal from a given set of samples. The signal is a finite length vector that is sparse in some given basis and is assumed to lie in a union of subspaces.
Aldroubi et al in \cite{ACM08} established the existence of an optimal union of subspaces model
for a given data set in an abstract setting and considered the finite dimensional case and the shift-invariant case for $L^2({\mathbb{R}}^d)$. They also developed an algorithm to find the model that fits the data set.

There are two technical aspects in the approach of Lu and Do that restrict the applicability  of their results in the shift-invariant space case.
The first one is due to the fact that the conditions are obtained in terms of Riesz bases
of translates of the SISs involved, and it is well known that not every SIS has a Riesz basis
of translates.
The second one is that the approach is based upon the sum of every two of the SISs in the union.
The conditions on the sampling operator are then obtained using fiberization  techniques on that sum.
This requires that the sum of each of two subspaces is a closed subspace, which is not true in general.

In this article we obtain the conditions for the sampling operator to be one-to-one and stable in terms
of {\it frames} of translates of the SISs instead of orthonormal basis. This extends the previous results to arbitrary SISs and in particular removes the restrictions mentioned above.
It is very important to have conditions based on frames, specially for applications, since frames are more flexible and simpler to construct. Frames of translates for shift-invariant spaces with generators that are smooth  and with good decay can be easily obtained.

On the other side we show that, using known results from the theory of SISs,  it is possible to determine families of subspaces on which the conditions for stability and injectivity are necessary and sufficient.

The article is organized in the following way:  Section 2 contains some notation and basic results that will be needed throughout. In Section 3 we set the problem of sampling in a union of subspaces in the general context of an abstract Hilbert space. We also give injectivity and stability conditions for the sampling operator, within this general setting. The case of finite-dimensional subspaces is studied in Section 4. In Section 5 we analyze the problem for   the Hilbert space  $L^2({\mathbb{R}}^n)$ and  sampling in a union of finitely generated shift-invariant spaces.
Finally in Section 6  we use the notion of angle between subspaces to obtain necessary and sufficient conditions
for the closedness of the sum of two shift-invariant spaces.

\section{Preliminaries}

We will assume that $I$ and $J$ are countable index sets and $\mathcal{H}$ is a separable Hilbert space
over the complex field.

\begin{definition}\label{frames}

A sequence $X=\{x_j\}_{j\in J}$  in $\mathcal{H}$ is a {\it Bessel sequence} if there
exists a constant $0<\beta<+\infty$ such that
$$\sum_{j\in J}|\langle h,x_j\rangle|^2\leq\beta\,\|h\|^2_{\mathcal{H}}\quad\forall\,h\in \mathcal{H}.$$

If in addition there exist constants $0<\alpha\leq\beta<+\infty$ such that

$$\alpha\,\|h\|^2_{\mathcal{H}}\leq\sum_{j\in J}|\langle h,x_j\rangle|^2\leq\beta\,\|h\|^2_{\mathcal{H}}\quad\forall\,h\in \mathcal{H},$$
then $X$ is said to be a {\it frame} for $\mathcal{H}$. The sequence X  is a {\it Parseval frame} if $\alpha$ and $\beta$ can be chosen so that $\alpha=\beta=1.$

The closure of the span of $X$ will be denoted by $\overline{\text{span}} \{x_j\}_{j\in J}$. $X$ is a
{\it frame sequence} if it is a frame for $\overline{\text{span}} \{x_j\}_{j\in J}$.

\end{definition}

\begin{definition}

A sequence $X=\{x_j\}_{j\in J}$ in $\mathcal{H}$ is a {\it Riesz basis} for $\mathcal{H}$ if it is complete in $\mathcal{H}$
and there exist constants $0<\alpha\leq\beta<+\infty$ such that

$$\alpha\,\sum_{j\in J}|c_j|^2\leq\Big\|\sum_{j\in J}c_j x_j\Big\|^2\leq\beta\,\sum_{j\in J}|c_j|^2\quad\forall\,\{c_j\}_{j\in J}\in \ell^2(J).$$

\end{definition}

\begin{definition}
If $X=\{x_j\}_{j\in J}$ is a Bessel sequence in $\mathcal{H}$, we define
the \textit{analysis operator} as
$$B_X:\mathcal{H}\rightarrow \ell^2(J),\quad B_Xh=\{\langle
h,x_j\rangle\}_{j\in J}.$$
The adjoint of $B$ is the \textit{synthesis operator}, given by
$$B_X^*:\ell^2(J)\rightarrow \mathcal{H},\quad B_X^*c=\sum_{j\in J}
c_j x_j.$$
The Bessel condition guarantees the boundedness of $B_X$ and as a consequence, that of $B_X^*$.
\end{definition}

\begin{definition}\label{Gramian}
Suppose $X=\{x_j\}_{j\in J}$ is a Bessel sequence in $\mathcal{H}$ and $B_X$ is the analysis operator.
The Gramian of the system $X$ is defined by
$$G_X:\ell^2(J)\rightarrow \ell^2(J),\quad G_X:=B_XB_X^*.$$
We identify  $G_{X}$ with its matrix representation.
$$(G_{X})_{j,k}=\langle x_k,x_j\rangle\quad\forall\,j,k\in J.$$
\end{definition}

Given a Hilbert space $\mathcal{L}$  and a bounded linear operator $W:\mathcal{L}\rightarrow\mathcal{L}$ , we will denote by $\sigma(W)$ the spectrum of $W$.

The following is a well known property. Its proof can be deduced from \cite[Lemma 5.5.4]{Chr03},

\begin{proposition}\label{fra}

Let $X:=\{x_j\}_{j\in J}\subseteq \mathcal{H}$ be a Bessel sequence, then $X$ is a frame sequence with constants $\alpha$ and $\beta$ if and only if
$$\sigma (G_X)\subseteq \{0\}\cup[\alpha,\beta].$$
\end{proposition}

\begin{definition}
Let $X:=\{x_j\}_{j\in J}$ and $Y:=\{y_i\}_{i\in I}$ be Bessel sequences in $\mathcal{H}$. Let  $B_{X}$ and $B_{Y}$ be the analysis operators associated to $X$ and $Y$ respectively. The {\it cross-correlation} operator is defined by
\begin{equation}\label{Gcross}
G_{X,Y}:\ell^2(J)\rightarrow \ell^2(I),\quad G_{X,Y}:= B_{Y}B_{X}^*.
\end{equation}
Identifying  again  $G_{X,Y}$ with its matrix representation, we write
$$(G_{X,Y})_{i,j}=\langle x_j,y_i\rangle\quad\forall\, j\in J,\forall\, i \in I.$$
\end{definition}

We will need the following property of Parseval frames.

\begin{proposition}\label{obpro1}

If $X =\{x_j\}_{j\in J}$ is a Parseval frame for a closed subspace $S$, and $B_X$ is the analysis operator associated to $X$,
then the orthogonal projection of $\mathcal{H}$ onto $S$ is
$$P_S=B_X^*B_X:\mathcal{H}\rightarrow\mathcal{H}, \quad P_Sh=\sum_{j\in J}\langle h, x_j\rangle\, x_j.$$
\end{proposition}

\begin{remark}
A Parseval frame for a Hilbert space does not need to be an orthogonal system. In fact, it is orthogonal if and only if every element of the set has unitary norm.
A simple example is the family  $X=\{\frac{1}{\sqrt{2}}e_1, \frac{1}{\sqrt{2}}e_1,e_n\}_{n\geq 2}$
where  $\{e_n\}_{n\in{\mathbb{N}}}$ is an orthonormal basis for the Hilbert space.
X is a Parseval frame that is not orthogonal and it is not even a basis.
\end{remark}

\section{The sampling operator}\label{section3}

Let $\mathcal{H}$ be a separable Hilbert space and $S \subseteq\mathcal{H}$ an arbitrary set.
Given $\Psi=\{\psi_i\}_{i\in I}$ a Bessel sequence in $\mathcal{H}$, the sampling problem consists of reconstructing
a signal $f\in S$ using the data $\{\langle f,\psi_i\rangle\}_{i\in I}$. We first require that the signals are uniquely determined by the data. That is, if we define the
{\it Sampling operator} by
\begin{equation}\label{op}
A:\mathcal{H}\rightarrow \ell^2(I), \quad Af:=\{\langle f,\psi_i\rangle\}_{i\in I},
\end{equation}
we require $A$ to be one-to-one on $S$. The set $\Psi$ will be called the {\it Sampling set}.

Note that the sampling operator $A$ is the analysis operator for the sequence  $\Psi$.

Another important property that is usually required for a sampling operator, is stability.
This is crucial to bound the error of  reconstruction in noisy situations.

The stable sampling condition was first proposed by \cite{Lan67} for the case when $S$ is the Paley-Wiener space.
It was then generalized in \cite{DL08} to the case when $S$ is a union of subspaces.

\begin{definition}
A sampling operator  $A$ is called {\it stable} on $S$ if there exist two constants
$0<\alpha\leq\beta<+\infty$ such that
$$\alpha\|x_1-x_2\|^2_{\mathcal{H}}\leq\|Ax_1-Ax_2\|^2_{\ell^2(I)}\leq\beta\|x_1-x_2\|^2_{\mathcal{H}}\quad\forall\,x_1,x_2\in S.$$
\end{definition}

When $S$ is a closed subspace, the injectivity and the stability can be expressed in terms of conditions on $P_S\Psi$, where $P_S$ is the orthogonal
projection of $\mathcal{H}$ onto $S$.

\begin{proposition}\label{reiny}
Let ${\mathcal{H}}$ be a Hilbert space,  $S \subseteq {\mathcal{H}}$  a closed subspace and $\Psi=\{\psi_i\}_{i\in I}$ a Bessel sequence in $\mathcal{H}$. If $A$ is the sampling operator associated to $\Psi,$ then we have
\begin{enumerate}
\item[i)] \label{one} The  operator $A$ is {\em one-to-one} on $S$ if and only if
$\{P_S\psi_i\}_{i\in I}$ is complete in S, that is $S=\overline{\textnormal{span}}\{P_S\psi_i\}_{i\in I}$.
\item [ii)]  The operator A is {\em stable} on $S$ with constants $\alpha$ and $\beta$ if and only if
$\{P_S\psi_i\}_{i\in I}$
is a frame for $S$ with constants $\alpha$ and $\beta$.
\end{enumerate}
\end{proposition}

\begin{proof}

The proof of {\it i)} is straightforward using that if $f\in S$ then
$$\langle f, P_S\psi_i\rangle=\langle P_S f,\psi_i\rangle=\langle f,\psi_i\rangle.$$
 For {\it ii)} note that for all  $f \in S$
$$\|Af\|^2_{\ell^2(I)}=\sum_{i\in I}|\langle f, \psi_n\rangle|^2=\sum_{i\in I}|\langle P_S f,\psi_n\rangle|^2=
\sum_{i\in I}|\langle f,P_S\psi_n\rangle|^2.$$
\end{proof}

\begin{remark}
Given a closed subspace $S$ in a Hilbert space ${\mathcal{H}}$, a sequence of vectors $\{\psi_i\}_{i\in I} \subseteq {\mathcal{H}}$
is called an {\it outer frame } for $S$ if $\{P_S\psi_i\}_{i \in I}$ is a frame for $S$. The notion of
outer frame was introduced in \cite{ACM04}. See also \cite{FW01} and \cite{LO04} for  related definitions.
Using this terminology, part ii) of Proposition \ref{reiny} says that the sampling operator $A$ is stable if
and only if  $\{\psi_i\}$ is an outer frame for $S$.
\end{remark}

In what follows we will extend one-to-one and stability conditions for the operator $A$,
to the case of a union of subspaces instead of a single subspace.

If $\{S_{\gamma}\}_{\gamma\in\Gamma}$ are closed subspaces of $\mathcal{H}$, with $\Gamma$ an arbitrary index set.
Let $$\chi:=\bigcup_{\gamma\in\Gamma} S_{\gamma}.$$

We want to study conditions on $\Psi$ so that the sampling operator $A$ defined by (\ref{op})
is one-to-one and stable on $\chi$.

This study continues the one initiated by Lu and Do \cite{DL08} in which they translated the conditions on $\chi$ into conditions on the subspaces defined by
\begin{equation}\label{suma}
S_{\gamma,\theta}:=S_{\gamma}+S_{\theta}=\{x+y\,:\,x\in S_{\gamma},\,y\in S_{\theta}\}.
\end{equation}

Working with the subspaces $S_{\gamma,\theta}$ instead of $\chi$, allows to exploit lineal pro\-perties of $A$.

They proved the following proposition.

\begin{proposition}\cite{DL08}\label{proposition-dl} With the above notation we have,
\begin{enumerate}
\item[i)]
The operator A is one-to-one on $\chi$ if and only if A is one-to-one on every  $S_{\gamma,\theta}$
with $\gamma,\theta\in\Gamma.$
\item[ii)]
The operator A is stable for $\chi$ with stability bounds $\alpha$ and $\beta$, if and only if A is stable for $S_{\gamma,\theta}$ with stability bounds $\alpha$ and $\beta$
for all $\gamma,\theta\in\Gamma,$ i.e.
$$\alpha\|x\|^2_{\mathcal{H}}\leq\|Ax\|^2_{\ell^2(I)}\leq\beta\|x\|^2_{\mathcal{H}}\quad\forall\,x\in S_{\gamma,\theta},\forall\ \gamma,\theta\in\Gamma.$$
\end{enumerate}
\end{proposition}

The sum of two closed infinite-dimensional  subspaces of a Hilbert space is not necessarily
 closed (see Example \ref{ejclau}). Furthermore, the injectivity of an operator on a subspace does not imply the injectivity on its closure. So,
we can not apply Proposition \ref{reiny} to the subspaces $S_{\gamma,\theta}.$
However, we can obtain a sufficient condition for the injectivity.
\begin{proposition}\label{complete}
 If $\{P_{\overline{S}_{\gamma,\theta}}\psi_i\}_{i\in I}$ is complete on $\overline{S}_{\gamma,\theta}$ for every $\gamma,\theta\in\Gamma,$ then $A$ is one-to-one on $\chi.$
\end{proposition}

When the  subspaces of the family $\{{S}_{\gamma,\theta}\}_{\gamma,\theta\in\Gamma}$ are all  closed,
the condition in Proposition \ref{complete}, will be also necessary for
the injectivity of $A$ on $\chi$.
So, a natural question will be, when the sum of two closed subspaces  of a Hilbert space is  closed.
In Section \ref{section6} we study this problem in several situations.

In the case of the stability, Proposition \ref{reiny} can be applied since, by the boundedness of $A$, we have the following.

\begin{proposition}\label{lemstab}
Let $S$ be a subspace of $\mathcal{H}$, the operator A is stable for $S$  with constants $\alpha$ and $\beta$ if and only if it is stable for
$\overline{S}$ with constants $\alpha$ and $\beta$.
\end{proposition}

As a consequence of this, using Propositions \ref{reiny} and  part ii) of Proposition \ref{proposition-dl}, we
have

\begin{proposition}
$A$ is stable for $\chi$ with constants
$\alpha$ and $\beta$ if and only if
$\{P_{\overline{S}_{\gamma,\theta}}\psi_i\}_{i\in I}$ is a frame for $\overline{S}_{\gamma,\theta}$ for every $\gamma,\theta\in\Gamma$ with the same constants $\alpha$ and $\beta$.
\end{proposition}

\section{Union of finite-dimensional subspaces}\label{section4}

In this section  we will first obtain conditions on the sequence $\{\psi_i\}_{i\in I}$ for the sampling operator to be  one-to-one on a union of finite-dimensional subspaces. We will then analyze the stability requirements.
We are interested in expressing these conditions in terms of the generators of the sum of every two subspaces of the union.

\subsection{The one-to-one condition for the sampling operator}

Let $\mathcal{H}$ be a Hilbert space, $\Psi=\{\psi_i\}_{i\in I}$  a Bessel sequence in $\mathcal{H}$,  and $A$ the sampling operator associated to $\Psi$ as in (\ref{op}).

Let $S$ be a finite-dimensional subspace of ${\mathcal{H}}$ and  $\Phi=\{\phi_j\}_{j=1}^{d}$ a finite frame for $S$.
(Recall that a finite set of vectors from a finite-dimensional subspace is a frame for that subspace if and only if it spans it.)

The cross-correlation operator associated to $\Psi$ and $\Phi$ (see (\ref{Gcross})) in this case can be written as,
$$G_{\Phi,\Psi}:{\mathbb{C}}^d\rightarrow \ell^2(I),\quad G_{\Phi,\Psi}=AB_{\Phi}^*,$$
where $B^*_{\Phi}:{\mathbb{C}}^d\rightarrow \mathcal{H}$ is the synthesis operator associated to $\Phi$.

The next theorem gives necessary and sufficient  conditions on the cross-correlation operator for the sampling operator to be one-to-one on $S$.

\begin{theorem}\label{teinys}
Let $\Psi=\{\psi_i\}_{i\in I}$ be a Bessel sequence for $\mathcal{H}$, $S$ a finite-dimensional  subspace of $\mathcal{H}$ and $\Phi=\{\phi_j\}_{j=1}^{d}$ a frame for $S.$
Then the following are equivalent:
\begin{enumerate}
\item [i)] $\Psi$ provides a one-to-one sampling operator on $S.$
\item [ii)] $\ker(G_{\Phi,\Psi})=\ker(B^*_{\Phi}).$
\item [iii)] $\dim(\textnormal{range}(G_{\Phi,\Psi}))=\dim(S).$
\end{enumerate}
\end{theorem}

\begin{proof}
The proof is straightforward using that the range of the operator $B^*_{\Phi}$ is $S$.
\end{proof}

\begin{remark}
Note that the conditions in Theorem \ref{teinys} do not depend on the particular  chosen frame.
That is, if there exists a frame $\Phi$ for $S$, such that
 $\dim(\text{range}(G_{\Phi,\Psi}))=\dim(S),$ then
  $\dim(\text{range}(G_{\widetilde{\Phi},\Psi}))=\dim(S),$ for any
 frame $\widetilde{\Phi}$ for $S$.
\end{remark}

Now  we will apply the previous theorem for the case of a union of subspaces.

Let $\{S_{\gamma}\}_{\gamma\in\Gamma}$ be a collection of finite-dimensional subspaces of $\mathcal{H}$, with $\Gamma$ an arbitrary index set.
Define, $$\chi:=\bigcup_{\gamma\in\Gamma} S_{\gamma}.$$
As before, set $S_{\gamma,\theta}:=S_{\gamma}+S_{\theta}.$

We obtain the following result which extends the result in \cite{DL08} to the case that the subspaces in the union are described by frames.

\begin{theorem}\label{teiny}
Let $\Psi=\{\psi_i\}_{i\in I}$ be a Bessel sequence for $\mathcal{H}$ and for every $\gamma,\theta\in\Gamma$, let $\Phi_{\gamma,\theta}$ be a frame for $S_{\gamma,\theta}$,
the following are equivalent:
\begin{enumerate}
\item [i)] $\Psi$ provides a one-to-one sampling operator on $\chi.$
\item [ii)] $\dim(\textnormal{range}(G_{\Phi_{\gamma,\theta},\Psi}))=\dim(S_{\gamma,\theta})$ for all $\gamma,\theta\in\Gamma.$
\end{enumerate}
\end{theorem}

Note that if $I$ is a finite set, the problem of testing the injectivity of $A$ on $\chi$ reduces to check that the rank of the cross-correlation
matrices are equal to the dimension of the subspaces $S_{\gamma,\theta}$.

In this case a lower bound for the cardinality of the sampling set can be established. This is stated
in the following corollary from \cite{DL08}. We include a proof of the result based on Theorem \ref{teiny}.

Here  $\# I$  denotes the cardinality of the finite set $I$.

\begin{corollary}
If the operator A is one-to-one on $\chi$ and $I$ is finite, then
$$\# I \geq \sup_{\gamma,\theta\in \Gamma}(\dim(S_{\gamma,\theta})).$$
\end{corollary}

\begin{proof}
Since $I$ is finite, we have that $\text{range} (G_{\Phi_{\gamma,\theta},\Psi})\subseteq{\mathbb{C}}^{\#I}$. Thus, using
part ii) of Theorem \ref{teiny}, we obtain that
$$\dim(S_{\gamma,\theta})=\dim(\text{range} (G_{\Phi_{\gamma,\theta},\Psi}))\leq \# I,\quad\forall\,\gamma,\theta\in\Gamma.$$
\end{proof}

\subsection{The stability condition for the sampling operator}

We are now interested in studying  conditions for stability of the sampling ope\-rator.
These conditions will be set in terms of the  cross-correlation operator.
We will consider Parseval frames to obtain simpler conditions.

Given Hilbert spaces $\mathcal{L}$ and $\mathcal{L'}$ and a bounded linear operator $W:\mathcal{L}\rightarrow\mathcal{L'}$, we denote by $\sigma^2(W)$  the set
$$\sigma^2(W)=\sigma(W^*W).$$

\begin{theorem}\label{teosta}
Let $\Psi=\{\psi_i\}_{i\in I}$ be a Bessel sequence for $\mathcal{H}$, $S$ a finite-dimensional  subspace of $\mathcal{H}$ and $\Phi$ a Parseval frame for $S$.

The sequence $\Psi$ provides a stable sampling operator for $S$  with constants $\alpha$ and $\beta$ if and only if
\begin{enumerate}
\item [i)] $\dim(\textnormal{range}(G_{\Phi,\Psi}))=\dim(S)$ and
\item [ii)] $\sigma^2(G_{\Phi,\Psi})\subseteq\{0\}\cup[\alpha,\beta].$
\end{enumerate}
\end{theorem}

\begin{proof}

Let $W:\mathcal{H}\rightarrow \ell^2(I),$ be  the analysis operator associated to
$P_S\Psi$.
For $x \in {\mathcal{H}}$, the equation,
$$Wx=\{\langle x,P_{S}\psi_i\rangle\}_{i\in I}=
\{\langle P_{S}x,\psi_i\rangle\}_{i\in I}=AP_{S}x,$$
shows that $W=AP_{S}.$

Since $\Phi$ is a Parseval frame for $S$, by Proposition \ref{obpro1}, $P_S = B_{\Phi}^*B_{\Phi}^{}$ then,
\begin{equation}\label{cuenta}
G_{P_S\Psi}=WW^*=
AP_{S}P_{S}A^*
=AP_{S}A^*=AB_{\Phi}^*B_{\Phi}^{}A^*=G_{\Phi,\Psi}G_{\Phi,\Psi}^*.
\end{equation}

Let us assume first that $A$ is stable for $S$. Item i) follows from Theorem \ref{teinys}.
Now we prove ii).

Since  $ S$ is closed ($S$ is finite dimensional) then Proposition \ref{reiny} gives that
$P_S\Psi:=\{P_{S}\psi_i\}_{i\in I}$
is a frame for $S$ with constants $\alpha$ and $\beta$. Using Proposition \ref{fra}, we have,
$$\sigma (G_{P_S\Psi})\subseteq \{0\}\cup[\alpha,\beta].$$

So, by (\ref{cuenta}),
$$\sigma (G_{P_S\Psi})=\sigma (G_{\Phi,\Psi}G_{\Phi,\Psi}^*)\subseteq \{0\}\cup[\alpha,\beta].$$
Finally, since (see  \cite{Rud91})
$$\sigma (G_{\Phi,\Psi}^*G_{\Phi,\Psi})\subseteq \{0\}\cup\sigma (G_{\Phi,\Psi}G_{\Phi,\Psi}^*),$$
it follows that
$$
\sigma ^2(G_{\Phi,\Psi})\subseteq\{0\}\cup[\alpha,\beta].$$

Suppose now that i) and ii) hold. Recall that A is stable for $S$ with stability bounds $\alpha,\beta$ if and only if $P_S\Psi:=\{P_{S}\psi_i\}_{i\in I}$ is
a frame for $S$ with frame bounds $\alpha,\beta$.

By Theorem \ref{teinys}, condition i) implies that the sampling operator is one-to-one on $S$.
Thus, using Proposition \ref{reiny},
$P_S\Psi:=\{P_{S}\psi_i\}_{i\in I}$ is complete in $S.$

That $P_S\Psi:=\{P_{S}\psi_i\}_{i\in I}$  is a frame sequence is straightforward by ii),
(\ref{cuenta}) and Proposition \ref{fra}.
\end{proof}

\begin{remark}
As in the case of injectivity, we note that the condition of stability does not depend on the  chosen Parseval frame. That means, if condition i) and ii) in the previous theorem hold for a Parseval frame $\Phi$ for $S$, then they hold for any Parseval frame $\widetilde{\Phi}$ for $S$.
\end{remark}

Theorem \ref{teosta} applied to the union of subspaces gives:

\begin{theorem}\label{teo}
Let $\Psi=\{\psi_i\}_{i\in I}$ be a set of sampling vectors and for every $\gamma,\theta\in\Gamma$, let $\Phi_{\gamma,\theta}$ be a Parseval frame for $S_{\gamma,\theta}$.

The sequence $\Psi$ provides a stable sampling operator for $\chi$  with constants $\alpha$ and $\beta$ if and only if
\begin{enumerate}
\item [i)] $\dim(\textnormal{range}(G_{\Phi_{\gamma,\theta},\Psi}))=\dim(S_{\gamma,\theta})$ for all $\gamma,\theta\in \Gamma$ and
\item [ii)] $\sigma^2(G_{\Phi_{\gamma,\theta},\Psi})\subseteq\{0\}\cup[\alpha,\beta]$  for all $\gamma,\theta\in\Gamma.$
\end{enumerate}
\end{theorem}

For examples and existence of sequences  $\Psi$ which verify the conditions of injectivity or stability in a union of finite-dimensional subspaces, we refer the reader to \cite{BD07} and \cite{DL08}.

\section{Sampling in a union of finitely generated shift-invariant spaces}

In this section we will consider the case of the Hilbert space   $\mathcal{H}=L^2({\mathbb{R}}^n)$ and
finitely generated shift-invariant spaces (FSISs). That is, we will study sampling in a union of FSISs.

We will first review some properties of these spaces. For a detailed treatment see  \cite{BDVR94,Bow00, Hel64, RS95} and the references therein.

\subsection{Some basic facts about shift-invariant spaces }

We use the Fourier transform defined by $$\hat{f}(\omega)=\int_{{\mathbb{R}}^n}f(x)\,e^{-2\pi i\omega x}\,dx$$
for $f\in L^1({\mathbb{R}}^n)$, and extended to be a unitary operator on $L^2({\mathbb{R}}^n)$.

$\mathbb{T}^n={\mathbb{R}}^n/{\mathbb{Z}}^n$ is the torus group, identified with $[0,1)^n$.

The translation by $k\in{\mathbb{Z}}^n$ is denoted by $t_k f:=f(\cdot-k).$

\begin{definition}
A closed subspace $S\subseteq L^2({\mathbb{R}}^n)$ is a {\it shift-invariant space} (SIS) if $f\in S$ implies $t_k f\in S$ for any $k\in{\mathbb{Z}}^n.$

Given $\Phi$ a set of functions in $L^2({\mathbb{R}}^n)$, we denote by $E(\Phi)$ the set,
$$E(\Phi):=\{t_k\phi:  k\in{\mathbb{Z}}^n, \phi \in \Phi \}.$$

The SIS generated by this set is $$S(\Phi):=\overline{\text{span}}(E(\Phi)).$$
If $S=S(\Phi)$ for some finite set $\Phi$ we say that $S$ is a \textit{finitely generated shift-invariant space} (FSIS).

If $S$ is an FSIS, we call  the {\it length} of S, the cardinality of a smallest generating
set for S, and write
$$\mbox{len}(S):=\min\{\#\Phi\,:\,S=S(\Phi)\}.$$
\end{definition}

Although the FSISs are infinite-dimensional subspaces, most of their pro\-perties can be translated into properties on the
fibers of the spanning sets. That allows to work with finite-dimensional subspaces of  $\ell^2({\mathbb{Z}}^n)$. We will
give the definition and some properties of the fibers.

The Hilbert space of square integrable vector functions $L^2(\mathbb{T}^n,\ell^2({\mathbb{Z}}^n))$,
consists of all vector valued measurable functions $F:\mathbb{T}^n\rightarrow \ell^2({\mathbb{Z}}^n)$ such that
$$\|F\|:=\Big(\int_{\mathbb{T}^n}\|F(x)\|^2_{\ell^2}\,dx\Big)^{\frac{1}{2}},$$is finite.

\begin{proposition}
The function $\tau:L^2({\mathbb{R}}^n)\rightarrow L^2(\mathbb{T}^n,\ell^2({\mathbb{Z}}^n))$ defined for $f\in L^2({\mathbb{R}}^n)$ by
$$\tau f(\omega):=\{\hat{f}(\omega+k)\}_{k\in{\mathbb{Z}}^n},$$
is an isometric isomorphism between  $L^2({\mathbb{R}}^n)$ and $L^2(\mathbb{T}^n,\ell^2({\mathbb{Z}}^n))$.

The sequence  $\{\hat{f}(\omega+k)\}_{k\in{\mathbb{Z}}^n}$  is called the {\it fiber} of $f$ at $\omega$.
\end{proposition}

\begin{definition}\label{def-range}
A \textit{range function} is a mapping
$$J:\mathbb{T}^n\rightarrow\{\text{closed subspaces of } \ell^2({\mathbb{Z}}^n)\}.$$
$J$ is measurable if the operator valued function of the orthogonal projections $\omega\mapsto P_{J(\omega)}$ is weakly measurable.

Note that in a separable Hilbert space measurability is equivalent to weak measurability. Therefore, the measurability of $J$
is equivalent to $\omega\mapsto P_{J(\omega)}(a)$ being vector measurable for each $a\in\ell^2({\mathbb{Z}}^n)$, or
$\omega\mapsto P_{J(\omega)}(F(\omega))$ being vector measurable for each fixed vector measurable function
$F:\mathbb{T}^n\rightarrow \ell^2({\mathbb{Z}}^n)$.
\end{definition}

\begin{proposition}\label{fun}
A closed subspace $S\subseteq L^2({\mathbb{R}}^n)$ is shift-invariant if and only if
$$
S=\{f\in L^2({\mathbb{R}}^n)\,:\,\tau f (\omega)\in J_S(\omega)\text{ for a.e. }\omega\in\mathbb{T}^n\},
$$
where $J_S$ is a measurable range function. The correspondence between $S$ and $J_S$ is one-to-one.

Moreover, if $S=S(\Phi)$ for some countable set $\Phi\subseteq L^2({\mathbb{R}}^n)$, then
$$J_S(\omega)=\overline{\textnormal{span}}\{\tau\phi (\omega)\,:\,\phi\in\Phi\}\quad\text{for a.e. }\omega\in\mathbb{T}^n.$$ The subspace $J_S(\omega)$ is called
 the \textit{fiber space} of $S$ at $\omega$.
\end{proposition}

\begin{proposition}\label{proy}
Let $S$ be a SIS of $L^2({\mathbb{R}}^n)$ and $f\in L^2({\mathbb{R}}^n),$ then
$$\tau(P_S f)(\omega)=P_{J_S(\omega)}(\tau f (\omega))\quad\mbox{ for a.e. }\omega\in\mathbb{T}^n.$$
\end{proposition}

\begin{definition}
Given $S$ a SIS of $L^2({\mathbb{R}}^n)$, the  {\it dimension function} associated to $S$ is defined by
$$\dim_S:\mathbb{T}^n\rightarrow{\mathbb{N}}_0\cup\{\infty\},\quad\dim_S(\omega)=\dim(J_S(\omega)).$$
\end{definition}
Here ${\mathbb{N}}_0$ denotes the set of non-negative integers.

The next two theorems characterize Bessel sequences, frames and Riesz bases of translates in terms of fibers.
\begin{theorem}\label{grami5}
Let $\Phi$ be a countable subset of $L^2({\mathbb{R}}^n)$. The following are equi\-valent.
\begin{enumerate}
\item[i)] $E(\Phi)$ is a Bessel sequence in $L^2({\mathbb{R}}^n)$ with constant $\beta.$
\item[ii)] $\tau\Phi (\omega):=\{\tau\phi (\omega)\,:\,\phi\in\Phi\}$ is a Bessel sequence in $\ell^2({\mathbb{Z}}^n)$ with constant $\beta$ for a.e. $\omega\in\mathbb{T}^n.$
\end{enumerate}
\end{theorem}

Let $\Phi = \{\phi_j\}_{j\in J}$ be a countable set of functions in $L^2({\mathbb{R}}^n)$ such that $E(\Phi)$ is a Bessel sequence. The Gramian of $\Phi$ at $\omega\in\mathbb{T}^n$ is
$$\mathcal{G}_{\Phi}(\omega):\ell^2(J)\rightarrow \ell^2(J), \quad
(\mathcal{G}_{\Phi}(\omega))_{i,j}=\langle \tau \phi_j(\omega),\tau\phi_i(\omega)\rangle_{\ell^2({\mathbb{Z}}^n)}\quad\forall\,i,j\in J.$$
Note that $\mathcal{G}_{\Phi}(\omega)$ is the Gramian operator associated to  the Bessel sequence $\{\tau\phi_j(\omega)\}_{j\in J}$ in $\ell^2({\mathbb{Z}}^n)$, that is  $\mathcal{G}_{\Phi}(\omega)= G_{\tau\Phi(\omega)},$ (see Definition \ref{Gramian}).

\begin{theorem}\label{grami3}
Let $S=S(\Phi)$, where $\Phi$ is a countable subset of $L^2({\mathbb{R}}^n)$. Then the following holds:
\begin{enumerate}
\item[i)] $E(\Phi)$ is a frame for $S$ with constants $\alpha$ and $\beta$ if and only if
$\tau\Phi(\omega)$ is a frame for $J_S(\omega)$ with constants $\alpha$ and $\beta$ for a.e. $\omega\in\mathbb{T}^n.$
\item[ii)] $E(\Phi)$ is a Riesz basis for $S$ with constants $\alpha$ and $\beta$ if and only if
$\tau\Phi(\omega)$ is a Riesz basis for $J_S(\omega)$ with constants $\alpha$ and $\beta$ for a.e. $\omega\in\mathbb{T}^n.$

Furthermore, if $\Phi$ is finite, $S$ has a Riesz basis of translates if and only if the dimension
function associated to $S$ is constant a.e. $\omega\in\mathbb{T}^n.$
\end{enumerate}
\end{theorem}

Let us remark here that if   $\Phi\subseteq L^2({\mathbb{R}}^n)$ is a set of generators for a shift-invariant space $S$, that is $S=S(\Phi)$,
then the set $E(\Phi)$ does not need to be a frame for $S$, even for finitely generated SISs. However it is always true that there exists a set  of generators for $S$ such that its integer translates form a frame for $S.$ This is the result of the next theorem.

\begin{theorem}\label{RS}
Given $S$ a SIS of $L^2({\mathbb{R}}^n)$, there exists a  subset $\Phi=\{\phi_j\}_{j\in J}\subseteq S$ such that
$E(\Phi)$ is a Parseval frame for $S$.
If S is finitely generated, the cardinal of $J$  can be chosen to be the length of $S.$
\end{theorem}

Although a SIS always has a frame of translates, there are SISs which have no Riesz bases of translates.
For example, consider the shift-invariant space $S$ generated by $\phi\in L^2({\mathbb{R}})$, where $\hat{\phi}(\omega)=\chi_{[0,\frac{1}{2})}(\omega).$
Since  $\dim_S(\omega)=1$ for a.e. $\omega\in[0,\frac{1}{2})$ and $\dim_S(\omega)=0$ for a.e. $\omega\in[\frac{1}{2},1)$, it follows by Theorem \ref{grami3} that $S$ has no Riesz bases of translates.

\subsection{Sampling from a Union of FSIS}

In this section we will study the sampling problem for the case in which the signal belongs to the set,
\begin{equation}\label{unionSIS}
\chi:=\bigcup_{\gamma\in\Gamma}S_{\gamma},
\end{equation}
where $S_{\gamma}$ are FSISs of $L^2({\mathbb{R}}^n)$.

In this setting, since our subspaces are shift-invariant, it is natural and also convenient that the sampling set will be the set of shifts from a fixed collection of functions in  $L^2({\mathbb{R}}^n)$, that is, the sampling operator will be given by a sequence
of integer translates of certain functions.

Given $\Psi:=\{\psi_i\}_{i\in I}$ such that   $E(\Psi)$ is a Bessel sequence in $L^2({\mathbb{R}}^n)$, we define the sampling operator associated to $E(\Psi)$ as
\begin{equation}\label{A}
A:L^2({\mathbb{R}}^n)\rightarrow \ell^2({\mathbb{Z}}^n\times I),\quad
Af=\{\langle f, t_k\psi_i\rangle\}_{i\in I, k\in{\mathbb{Z}}^n}.
\end{equation}
As we showed in Section \ref{section3} the conditions on  the sampling operator to be  one-to-one and stable in a union of subspaces can be established in terms  of one-to-one  and stability conditions on the sum of every two of the subspaces from the union.

However the condition that we have for the sampling operator to be one-to-one on a subspace, requires that the subspace is closed (Proposition \ref{reiny}).

Since the sum of two FSISs  is  not necessarily a closed subspace, the conditions should be imposed on the closure of the sum.

Conditions that guarantee  that the sum of two FSISs is closed are described in Section \ref{section6}.

In what follows we will consider, for each $\gamma,\theta \in \Gamma$, the subspaces,
\begin{equation}\label{defclau}
\overline{S}_{\gamma,\theta}:=\overline{S_{\gamma}+S_{\theta}}.
\end{equation}
The following proposition states that the closure of the sum of two SISs is a SIS generated by the union
of the generators of the two spaces. Its proof is straightforward.

\begin{proposition}\label{clau}
Let $\Phi$ and $\Phi'$ be sets in $L^2({\mathbb{R}}^n)$,
then $$\overline{S(\Phi)+S(\Phi') }=S(\Phi \cup\Phi').$$
In particular, if $S$ and $S'$are FSISs, then $\overline{S+S'}$ is an FSIS and
$$\textnormal{len}(\overline{S+S'}) \leq \textnormal{len}(S)+\textnormal{len}(S').$$
\end{proposition}

Now, as a consequence of Proposition \ref{clau}, for each $\gamma,\theta\in\Gamma$, $\overline{S}_{\gamma,\theta}$ is an FSIS.
Then, by Theorem \ref{RS}, we can choose, for each $\gamma,\theta\in\Gamma$, a {\it finite} set
$$\Phi_{\gamma,\theta}=\{\phi_j^{\gamma,\theta}\}_{ j=1}^{d_{\gamma,\theta}} $$
of $ L^2({\mathbb{R}}^n)$ functions such that,
$$\overline{S}_{\gamma,\theta}=S(\Phi_{\gamma,\theta}),$$ and
$E(\Phi_{\gamma,\theta})$ forms a Parseval frame for $\overline{S}_{\gamma,\theta}$.

\subsection{The one-to-one condition}

We now study the conditions that the sampling set must satisfy  in order  for the operator $A$ defined by (\ref{A}) to be one-to-one on $\chi$.

Given a shift-invariant space $S$, the orthogonal projection onto $S$, denoted by $P_S$, commutes
with integer translates. Then, part i) of  Proposition \ref{reiny} can be rewritten as,

\begin{proposition}\label{reb}
Given a shift-invariant space $S$, $\Psi=\{\psi_i\}_{i\in I}$  such that $E(\Psi)$ is a Bessel sequence in $L^2({\mathbb{R}}^n)$ and $A$ the  sampling operator associated to $E(\Psi)$. Then the following are equivalent.
\begin{enumerate}
\item [i)] The sampling operator $A$  is one-to-one on  $S.$
\item [ii)] $E( P_S\Psi)=\{t_k P_S\psi_i\}_{i\in I,k\in{\mathbb{Z}}^n}$
is complete in $S$, that is $S=\overline{\textnormal{span}} \,E( P_S\Psi)$.
\end{enumerate}
\end{proposition}

Since $E(\Psi)$  is a Bessel sequence in $L^2({\mathbb{R}}^n)$, by Theorem \ref{grami5} we have that
$\{\tau\psi_i(\omega)\}_{i\in I}$ is a Bessel sequence in $\ell^2({\mathbb{Z}}^n)$ for a.e $\omega\in\mathbb{T}^n$, so we can define (up to a set of measure zero), for $\omega \in \mathbb{T}^n$, the sampling operator related to the fibers:
$$\mathcal{A}(\omega):\ell^2({\mathbb{Z}}^n)\rightarrow \ell^2(I),$$ with
\begin{equation}\label{aw}
\mathcal{A}(\omega)(c)=\{\langle c, \tau \psi_i(\omega)\rangle\}_{i\in I}.
\end{equation}

That is, for a fixed $\omega \in \mathbb{T}^n$, we consider the problem of sampling from a union of subspaces in a different setting. The Hilbert space is $\ell^2({\mathbb{Z}}^n)$,  the sequences of the sampling set are $\{\tau\psi_i(\omega)\}_{i\in I}$, and the subspaces in the union are $J_{S_{\gamma}}(\omega),  \gamma \in \Gamma.$

Since the subspaces $\overline{S}_{\gamma,\theta}$ are FSISs, the fiber spaces $J_{\overline{S}_{\gamma\,\theta}}(\omega)$ are finite-dimensional.
So, the results of Section \ref{section4} can be applied, and conditions on the fibers can be obtained
in order for the operator $\mathcal{A}(\omega)$ to be one-to-one.

We are now going to show that given a finitely generated shift-invariant space $S$, the operator $A$ is one-to-one on $S$ if and
only if for almost every $\omega \in \mathbb{T}^n$,  the operator $\mathcal{A}(\omega)$ is one-to-one on the corresponding fiber spaces $J_{S}(\omega)$ associated to $S$.
Once this is accomplished, we can apply the known conditions for the operator $\mathcal{A}(\omega).$

Given $\{t_k\phi_j\}_{j=1,k\in{\mathbb{Z}}^n}^{d}$ a Bessel sequence in $L^2({\mathbb{R}}^n)$, we have the synthesis operator related to the fibers,
that is
\begin{equation}\label{b*w}
\mathcal{B}_{\Phi}^*(\omega):{\mathbb{C}}^{d}\rightarrow
\ell^2({\mathbb{Z}}^n),\quad
\mathcal{B}_{\Phi}^*(\omega)(c_1,\ldots,c_d)=\sum_{j=1}^{d}c_j\tau\phi_j(\omega).
\end{equation}
Note  that $\mathcal{B}_{\Phi}^*(\omega)$ is the synthesis operator associated to the set $\tau\Phi(\omega)$, that is $\mathcal{B}_{\Phi}^*(\omega)=B_{\tau\Phi(\omega)}^*$.

And we will have the cross-correlation operator associated to the fibers
$$\mathcal{G}_{\Phi,\Psi}(\omega):{\mathbb{C}}^d\rightarrow \ell^2(I),\quad \mathcal{G}_{\Phi,\Psi}(\omega):=\mathcal{A}(\omega)\mathcal{B}_{\Phi}^*(\omega),$$
\begin{equation}\label{gw}
(\mathcal{G}_{\Phi,\Psi}(\omega))_{i,j}=\langle \tau\phi_j(\omega),\tau\psi_i(\omega)\rangle\quad\forall\,1\leq j\leq d,i\in I.
\end{equation}
Again we should remark  that $\mathcal{G}_{\Phi,\Psi}(\omega)$ is the cross-correlation operator associated to $\tau\Phi(\omega)$ and $\tau\Psi(\omega)$, that is $\mathcal{G}_{\Phi,\Psi}(\omega)=G_{\tau\Phi(\omega),\tau\Psi(\omega)}$.

\begin{theorem}\label{teinysis}
Let $\Psi=\{\psi_i\}_{i\in I}$ be such that $E(\Psi)$ is a Bessel sequence in $L^2({\mathbb{R}}^n)$, $S$ an FSIS generated by a finite set $\Phi$, and $A$ the sampling operator associated to $E(\Psi)$, then
the following are equivalent:
\begin{enumerate}
\item [i)] $\Psi$ provides a one-to-one sampling operator for $S.$
\item [ii)] $\ker(\mathcal{G}_{\Phi,\Psi}(\omega))=\ker(\mathcal{B}_{\Phi}^*(\omega))$ for a.e. $\omega \in \mathbb{T}^n.$
\item [iii)] $\dim(\textnormal{range}(\mathcal{G}_{\Phi,\Psi}(\omega)))=\dim_{S}(\omega)$ for a.e. $\omega \in \mathbb{T}^n.$
\end{enumerate}
\end{theorem}

For the proof of Theorem \ref{teinysis} we need the following.
\begin{lemma}\label{leinysis}
Let $S$ be an FSIS$, \Psi=\{\psi_i\}_{i\in I}$  such that $E(\Psi)$ is a Bessel sequence in $L^2({\mathbb{R}}^n)$,  and $A$ the sampling operator associated to $E(\Psi)$.
Then $A$ is one-to-one on $S$ if and only if  $\mathcal{A}(\omega)$ is one-to-one on $J_S(\omega)$ for a.e. $\omega\in\mathbb{T}^n.$
\end{lemma}
\begin{proof}
Since S is a SIS, by Proposition \ref{reb}, $A$ is one-to-one on $S$ if and only if
\begin{equation}\label{aa}
S=\overline{\text{span}}\,E(P_S\Psi).
\end{equation}

By Proposition \ref{fun}, equation (\ref{aa}) is equivalent to
\begin{equation}\label{L2}
J_S(\omega)=\overline{\text{span}}\{\tau (P_S\psi_i)(\omega): i\in I\}\quad\mbox{ for a.e. }\omega\in\mathbb{T}^n.
\end{equation}
So, we have proved that  $A$ is one-to-one on $S$ if and only if
(\ref{L2}) holds.

On the other side, given $\omega\in\mathbb{T}^n$, and using Proposition \ref{reiny} for the sampling operator $\mathcal{A}(\omega)$
 and the space $\mathcal{H}=\ell^2({\mathbb{Z}}^n)$, we have that $\mathcal{A}(\omega)$ is one-to-one on
$J_S(\omega)$ if and only if
$$J_S(\omega)=\overline{\text{span}}\{ P_{J_S(\omega)}(\tau\psi_i(\omega)): i\in I\}.$$
Then, using Proposition \ref{proy}, we conclude that (\ref{L2}) holds if and only if $\mathcal{A}(\omega)$ is one-to-one on
$J_S(\omega)$, for a.e. $\omega\in\mathbb{T}^n$, which completes the proof of the lemma.

\end{proof}

\begin{proof}[{\it Proof of Theorem \ref{teinysis}}]
Since $\Phi$ is a set of generators for  $S$,  we have that
for a.e. $\omega\in \mathbb{T}^n$, $\tau\Phi(\omega)$ is a set of generators for  $J_S(\omega)$.

Now, for a.e. $\omega\in \mathbb{T}^n$ we can apply Theorem \ref{teinys} for the sampling operator $\mathcal{A}(\omega)$
and the finite-dimentional subspace $J_S(\omega)$ to obtain the equivalence of the following propositions:

\begin{enumerate}
\item [a)] $\mathcal{A}(\omega)$ is one-to-one on
$J_S(\omega).$
\item [b)] $\ker(\mathcal{G}_{\Phi,\Psi}(\omega))=\ker(\mathcal{B}_{\Phi}^*(\omega)).$
\item [c)] $\dim(\text{range}(\mathcal{G}_{\Phi,\Psi}(\omega)))=\dim(J_{S}(\omega))=\dim_{S}(\omega).$
\end{enumerate}
From here the proof follows using Lemma \ref{leinysis}.

\end{proof}

Note that with the previous theorem we have conditions for $A$ to be one-to-one on $\overline{S}_{\gamma,\theta}$,
and since
$$S_{\gamma,\theta}=S_{\gamma}+S_{\theta}\subseteq \overline{S}_{\gamma,\theta},$$
we obtain the following corollary.

\begin{corollary}\label{cor-iny-sis}
Let $E(\Psi)$ be a Bessel sequence in $L^2({\mathbb{R}}^n)$ for some set of functions $\Psi$. For every $\gamma,\theta\in\Gamma$, let $\Phi_{\gamma,\theta}$ be a finite set of generators for $\overline{S}_{\gamma,\theta}$.
If for each $\gamma,\theta\in\Gamma,$
$$
\dim(\textnormal{range}(\mathcal{G}_{\Phi_{\gamma,\theta},\Psi}(\omega)))=\dim_{\overline{S}_{\gamma,\theta}}(\omega)\quad\text{for a.e. }\omega \in \mathbb{T}^n,
$$ then $A$ is one-to-one on $\chi$.
\end{corollary}

\begin{remark}
It is important to note that the injectivity of $A$ on
$S_{\gamma,\theta}$ does not imply the injectivity on $\overline{S}_{\gamma,\theta}$, thus, we have only obtained sufficient
conditions for $A$ to be one-to-one.
This is not a problem in general, because  as we will see in the next section,
stability implies injectivity  in the case of  the sampling operator and stability
is a common and needed assumption in most  sampling applications.
\end{remark}

\subsection{The stability condition}

As a consequence of Proposition \ref{lemstab}, we will obtain
necessary and sufficient conditions for the stability of $A$.

As in the previous subsection, using that the orthogonal projection onto a SIS commutes with integer translates, we have the following
version of Proposition \ref{reiny}.

\begin{proposition}\label{2}
Given S a SIS of $L^2({\mathbb{R}}^n)$, $\Psi=\{\psi_i\}_{i\in I}$ such that $E(\Psi)$ is a Bessel sequence in $L^2({\mathbb{R}}^n)$ and $A$ the  sampling operator associated to $E(\Psi)$. Then the following are equivalent:
\begin{enumerate}
\item [i)] The sampling operator $A$ is stable for  $S$ with constants $\alpha$ and $\beta$.
\item [ii)] $E(P_S\Psi)$ is a frame for $S$ with constants $\alpha$ and $\beta$.
\end{enumerate}
\end{proposition}

Now we are able to state the stability theorem. We will use the operator related to the fibers, defined by
(\ref{aw}), (\ref{b*w}) and (\ref{gw}).

\begin{theorem}\label{stasub}
Let $\Psi=\{\psi_i\}_{i\in I}$ be such that $E(\Psi)$ is a Bessel sequence for $L^2({\mathbb{R}}^n)$ and $A$ the  sampling operator associated to $E(\Psi)$.
Let $S$ be an FSIS, and  $\Phi$ a finite set of functions such that $E(\Phi)$ forms a Parseval frame for $S.$

Then
$E(\Psi)$ provides a stable sampling operator for $S$
if and only if
\begin{enumerate}
\item [i)] $\dim(\textnormal{range}(\mathcal{G}_{\Phi,\Psi}(\omega)))=\dim_{S}(\omega)$ for a.e. $\omega\in \mathbb{T}^n$ and
\item [ii)] There exist constants $0<\alpha\leq\beta<\infty$ such that $$\sigma^2(
\mathcal{G}_{\Phi,\Psi}(\omega))\subseteq\{0\}\cup[\alpha,\beta]\quad\mbox{for a.e. }
\omega \in \mathbb{T}^n.$$
\end{enumerate}
\end{theorem}

\begin{proof}
$\Phi$ is a Parseval frame for $S$, so, by Theorem \ref{grami3}, we have that
 for a.e. $\omega\in \mathbb{T}^n$, $\tau\Phi(\omega)$ is a Parseval frame for $J_S(\omega)$. Since
 $J_S(\omega)$ is a finite-dimensional space of $\ell^2({\mathbb{Z}}^n)$, Theorem
\ref{teosta} holds for $\mathcal{A}(\omega)$.

So, we only have to prove that A is stable for $S$  with constants $\alpha$ and $\beta$ if and only if $\mathcal{A}(\omega)$ is stable for
$J_{S}(\omega)$
with constants $\alpha$ and $\beta$.

By Proposition \ref{2}, the stability of $A$ in $S$ is equivalent to
$E(P_{S}\Psi)$ being a frame for $S$ with constants $\alpha$ and $\beta$.
By Theorem \ref{grami3}, this is equivalent to $$\{\tau(P_{S} \psi _i)(\omega)\}_{i\in I}$$ being a frame for $J_{S}(\omega)$ with constants $\alpha$ and $\beta$ for a.e. $\omega\in \mathbb{T}^n.$

On the other hand, given $\omega\in \mathbb{T}^n$, the operator $\mathcal{A}(\omega)$ is stable for
$J_{S}(\omega)$, if and only if
$$\{P_{J_{S}(\omega)}(\tau\psi_i(\omega))\}_{i\in I}$$ is a frame for $J_{S}(\omega)$ with constants $\alpha$ and $\beta$.

The proof can be finished now using  first Proposition \ref{proy}, i.e.
$$\tau (P_{S}\psi_i)(\omega)=P_{J_{S}(\omega)}(\tau\psi_i(\omega))\quad\mbox{ for a.e. } \omega\in\mathbb{T}^n,$$
and then Theorem \ref{teosta}.
\end{proof}

Now we apply  Theorem \ref{stasub} and Proposition \ref{lemstab} to obtain the following.

\begin{theorem}\label{stasis}
Let $\Psi=\{\psi_i\}_{i\in I}$ such that $E(\Psi)$ is a Bessel sequence for $L^2({\mathbb{R}}^n)$, and for every $\gamma,\theta\in\Gamma$ let $\Phi_{\gamma,\theta}$ be a Parseval
frame for $\overline{S}_{\gamma,\theta}$. Then
$E(\Psi)$ provides a stable sampling operator for $\chi$
if and only if
\begin{enumerate}
\item [i)] $\dim(\textnormal{range}(\mathcal{G}_{\Phi_{\gamma,\theta},\Psi}(\omega)))=\dim_{\overline{S}_{\gamma,\theta}}(\omega)$ for a.e. $\omega\in \mathbb{T}^n,\,\forall\,
\gamma,\theta\in\Gamma$ and
 \item[ii)]There exist constants $0<\alpha\leq\beta<\infty$ such that $$\sigma^2(
\mathcal{G}_{\Phi_{\gamma,\theta},\Psi}(\omega))\subseteq\{0\}\cup[\alpha,\beta]\quad\mbox{for a.e. }
\omega \in \mathbb{T}^n,\,\forall\,
\gamma,\theta\in\Gamma.$$
\end{enumerate}
\end{theorem}

Finally, as in \cite{DL08}, we  obtain a lower bound for the amount of samples. In contrast to the previous section, we only find bounds for stable
operators. We can not say anything about one-to-one operators since we only obtained sufficient conditions for the injectivity.

\begin{proposition}
If the operator A is stable for $\chi$ and $I$ is finite, then
$$\# I\geq \sup_{\gamma,\theta\in \Gamma}(\textnormal{len}(\overline{S}_{\gamma,\theta})).$$
\end{proposition}

\begin{proof}
Since $I$ is finite, it holds that $\textnormal{range}(\mathcal{G}_{\Phi_{\gamma,\theta},\Psi}(\omega)))\subseteq {\mathbb{C}}^{\#I}$ for a.e.
$\omega\in \mathbb{T}^n$.
Hence, by Theorem \ref{stasis}, we have that
$$\dim_{\overline{S}_{\gamma,\theta}}(\omega)=\dim(\text{range}(\mathcal{G}_{\Phi_{\gamma,\theta},\Psi}(\omega)))\leq \#I\quad\mbox{ for a.e. }\omega\in \mathbb{T}^n,\,\forall\,
\gamma,\theta\in\Gamma.$$
This shows that, given $\gamma,\theta\in\Gamma,$
$$\mbox{ess-sup }\{\dim_{\overline{S}_{\gamma,\theta}}(\omega)\,:\,\omega\in\mathbb{T}^n\}\leq\# I.$$
The proof of the proposition follows using \cite[Theorem 3.5]{BDVR94}.

\end{proof}

We would like to note that based in our results, it is possible to state conditions for the injectivity and stability for the sampling operator
in a union of SISs which are not necessarily finitely-generated. For this, condition iii) of Theorem \ref{teinysis}
should be replaced by condition ii).

\section{Necessary and sufficient conditions for the closedness of the sum of two SIS.}\label{section6}

In this section we review the conditions for the sum of two closed subspaces of a Hilbert space to
be closed. Then we apply the results to the class of FSISs in $L^2({\mathbb{R}}^n)$.

\subsection{Closedness of the sum of two subspaces of a Hilbert space.}

Throughout this section $\mathcal{H}$ will be a separable Hilbert space.\\
If $S$ is a closed subspace of $\mathcal{H}$, we write $P_S$ for the orthogonal projection onto $S$.
Given $U$ and $V$ closed subspaces
of $\mathcal{H}$, we will use the symbol $P_U\big\vert_V$ to denote the restriction of $P_U$ to the subspace $V$.\\
The orthogonal complement of $U\cap V$ in $U$ will be denoted by
\begin{equation}\label{utilde}
U\ominus V:=U\cap(U\cap V)^{\perp}.
\end{equation}
If $A:\mathcal{H}\rightarrow \mathcal{H}$ is a bounded linear operator, we will use the norm
$$\|A\|=\sup_{x\neq 0}\frac{\|Ax\|}{\|x\|}.$$

The conditions on the closedness of the sum of two closed subspaces of $\mathcal{H}$ will be given in terms of the angle between the subspaces. We refer the reader to \cite{Deu95} for details and proofs.

\begin{definition}
Let $U$ and $V$ be closed subspaces of $\mathcal{H}$.
\begin{enumerate}
\item [a)] The  \textit{minimal angle}  between $U$ and $V$ (or \textit{Dixmier angle}) is the angle in $[0,\frac{\pi}{2}]$ whose cosine is $$\textbf{c}_0[U,V]:=\sup \{|\langle u,v\rangle|\,:\,u\in U, v\in V, \|u\|\leq 1, \|v\|\leq1\}.$$
\item [b)] The \textit{angle} between $U$ and $V$ (or \textit{Friedrichs angle}) is the angle in $[0,\frac{\pi}{2}]$ whose cosine is $$\textbf{c}[U,V]:=\sup\{|\langle u,v \rangle|\,:\,u\in
U\ominus V, v\in V\ominus U\text{ and }\|u\|\leq 1, \|v\|\leq 1\}.$$
\end{enumerate}
\end{definition}

Now we  state some known results concerning both notions of angles between subspaces.

\begin{proposition}\label{cerrados1}
Let $U$ and $V$ be closed subspaces of $\mathcal{H}$.
\begin{enumerate}
\item [i)]$\textbf{c}_0[U,V]=\|P_U\big\vert_V\|.$
\item[ii)]  $\textbf{c}[U,V]=\textbf{c}_0[U\ominus V,V\ominus U].$
\item [iii)] $U+V$ is closed if and only if $\textbf{c}[U,V]<1.$
\end{enumerate}
\end{proposition}

Let $\mathcal{H}$ and $\mathcal{K}$ be separable Hilbert spaces, and $T:\mathcal{H}\rightarrow \mathcal{K}$ a bounded linear operator with closed range. We denote by  $T^{\dagger}$,  the pseudo-inverse of $T$ (see \cite{Chr03} for definition and properties).

The following theorem provides a formula which gives an easy way of calculating the angle between two subspaces using operators associated to frames. Its proof follows from  \cite[Theorem 2.1]{KLL06} and part ii) of Proposition \ref{cerrados1}.

\begin{theorem}\label{teo-c}
Let $U$ and $V$ be closed subspaces of $\mathcal{H}$. Suppose that $X$ and $X'$ are countable subsets of $\mathcal{H}$ which are frames for $U\ominus V$ and $V\ominus U$ respectively. Then,
$$\textbf{c}[U,V]=\|(G_{X'}^{\dagger})^{\frac{1}{2}}G_{X,X'}(G_{X}^{\dagger})^{\frac{1}{2}}\|,$$
where $G_X$ and $G_{X'}$ are the Gramian operators and $G_{X,X'}$ is the cross-correlation operator.
\end{theorem}

\subsection{Closedness of the sum of two shift-invariant subspaces}

As it was stated in Proposition \ref{cerrados1}, the closedness of the sum of two subspaces depends on the Friedrichs angle between them.
In this section, we  provide an expression  for the Friedrichs angle between  two SISs in terms of the gramians of the generators.
In \cite{KLL06} Kim et al has found a similar expression for the Dixmier angle between two SISs.

The main theorem of this part gives necessary and sufficient conditions for the sum of two SISs to be closed. We first state the theorem and then we apply this result to obtain a more general version of Corollary \ref{cor-iny-sis}. The proof of the theorem will be given at the end of the section.

\begin{theorem}\label{teofinal}
Let $U$ and $V$ be SISs of $L^2({\mathbb{R}}^n)$. Suppose that $\Phi, \Phi'$ are sets of functions in $L^2({\mathbb{R}}^n)$ such that for a.e. $\omega\in\mathbb{T}^n$, $\tau\Phi(\omega)$ and $\tau\Phi' (\omega)$ are frames for $J_{U\ominus V}(\omega)$ and
 $J_{V\ominus U}(\omega)$ respectively.
Then, $U+V$ is closed if and only if
\begin{equation}\label{ec10}
\textbf{c}[U,V]=\textnormal{ess-sup } \{\|(\mathcal{G}_{\Phi'}(\omega)^{\dagger})^{\frac{1}{2}}\mathcal{G}_{\Phi,\Phi' }(\omega)(\mathcal{G}_{\Phi}(\omega)^{\dagger})^{\frac{1}{2}}\|\,:\, \omega\in\mathbb{T}^n\}<1.
\end{equation}
\end{theorem}

 Note that, if $S=S(\Phi)$ is an FSIS, we have that $\tau\Phi(\omega)$ is a frame for $J_S(\omega)$  for a.e. $\omega\in\mathbb{T}^n$, even though $E(\Phi)$ is not a frame for $S$. Thus, if $U$ and $V$ are FSISs,  condition (\ref{ec10}) can
 be checked on any set of generators of the subspaces $U\ominus V$ and $V\ominus U$. At the end of the section we give an example in which we compute the Friedrichs angle between two FSISs.

In the next theorem we show that imposing certain restrictions on the angle between the subspaces, we obtain
necessary and sufficient conditions for the injectivity of the sampling operator. This gives a more complete version of Corollary \ref{cor-iny-sis} .

\begin{theorem}
Let $\Psi=\{\psi_i\}_{i\in I}$ be such that $E(\Psi)$ is a Bessel sequence in $L^2({\mathbb{R}}^n)$ and suppose condition (\ref{ec10}) is satisfied for every $\gamma,\theta\in\Gamma$. If $\Phi_{\gamma,\theta}$ is a finite set of generators  for $S_{\gamma,\theta}$,
the following are equivalent:
\begin{enumerate}
\item [i)]$\Psi$ provides a one-to-one sampling operator for $\chi.$
\item [ii)]$\dim(\textnormal{range}(\mathcal{G}_{\Phi_{\gamma,\theta},\Psi}(\omega)))=\dim_{S_{\gamma,\theta}}(\omega)$ for a.e. $\omega \in \mathbb{T}^n, \forall\,\gamma,\theta\in\Gamma.$
\end{enumerate}
\end{theorem}

\begin{proof}
Since condition (\ref{ec10}) is satisfied for every $\gamma,\theta\in\Gamma$, it holds that the subspaces $S_{\gamma,\theta}$
are FSISs. The proof of the theorem follows applying Theorem \ref{teinysis} to these subspaces.

\end{proof}

In what follows we will give some lemmas which will be needed for the proof of Theorem \ref{teofinal}.
The results in these lemmas  are interesting by themselves.

The first lemma uses the notion of range function introduced in Definition \ref{def-range}.

\begin{lemma}\label{lema-rango-int}
Given $U$ and $V$ SISs  of $L^2({\mathbb{R}}^n)$. Then the range function
$$
R:\mathbb{T}^n\rightarrow\{\text{closed subspaces of } \ell^2({\mathbb{Z}}^n)\},\quad R(\omega)=J_U(\omega)\cap J_V(\omega),
$$ is measurable.
\end{lemma}

\begin{proof}\label{proy-int-med}
Recall that the measurability of $R$ is equivalent to $\omega\mapsto P_{J_U(\omega)\cap J_V(\omega)}$ being measurable.

It is known (see \cite{Neu50}) that given $M$ and $N$ closed subspaces of a separable Hilbert space $\mathcal{H}$, for each $x\in\mathcal{H}$,
$$
P_{M\cap N}(x)=\lim_{n\to +\infty}(P_MP_N)^n(x).
$$
Note that if we have two measurable functions
 $$Q_1,Q_2: \mathbb{T}^n \rightarrow \{\text{orthogonal projections in } \ell^2({\mathbb{Z}}^n)\},$$
 then the map $ \omega \mapsto Q_1(\omega)Q_2(\omega)$ is measurable.
 For, let $F$ be an arbitrary measurable function from $\mathbb{T}^n$ into $\ell^2({\mathbb{Z}}^n)$.
 Then $$ Q_1(\omega)Q_2(\omega)(F(\omega)) = Q_1(\omega)(Q_2(\omega)(F(\omega))).$$
 By Definition \ref{def-range}, the measurability of $Q_2(\omega)$ implies the vector measurability of $Q_2(\omega)(F(\omega)).$
Since $Q_1(\omega)$ is measurable, $ Q_1(\omega)Q_2(\omega)(F(\omega))$ is measurable. What shows that
$ \omega \mapsto Q_1(\omega)Q_2(\omega)$ is measurable.

As a consequence, it holds that for any $n \in {\mathbb{N}}$ the map $\omega\mapsto (P_{J_U(\omega)}P_{J_V(\omega)})^n$ is measurable, that is, for each $a\in\ell^2({\mathbb{Z}}^n)$, $\omega\mapsto (P_{J_U(\omega)}P_{J_V(\omega)})^n(a)$ is measurable.
From here the proof follows using that,
$$
P_{J_U(\omega)\cap J_V(\omega)}(a)=\lim_{n\to +\infty}(P_{J_U(\omega)}P_{J_V(\omega)})^n(a).
$$
\end{proof}

With the previous lemma we obtain the following property of the fiber spaces.

\begin{lemma}\label{Js}
Let $U$ and $V$ be SISs of $L^2({\mathbb{R}}^n)$. Then, $$J_{U\ominus V}(\omega)=J_U(\omega)\ominus J_V(\omega)\quad\text{ for a.e. }\omega\in \mathbb{T}^n.$$
\end{lemma}

\begin{proof}
We will first prove that
\begin{equation}\label{J-int}
J_{U\cap V}(\omega)=J_U(\omega)\cap J_V(\omega)\quad\text{ for a.e. }\omega\in \mathbb{T}^n.
\end{equation}
Let $R$ be the measurable range function defined in Lemma \ref{lema-rango-int}.
Since
$$
U\cap V=\{f\in L^2({\mathbb{R}}^n)\,:\,\tau f(\omega)\in R(\omega)\text{ for a.e. }\omega \in\mathbb{T}^n\},
$$
it follows that $R$ is the range function associated to the shift-invariant space $U\cap V$, thus
(\ref{J-int}) holds.

Using (\ref{J-int}), the proof of the proposition is straightforward as
$$(J_S(\omega))^{\perp}=J_{S^{\perp}}(\omega)\quad\text{for a.e. }\omega\in \mathbb{T}^n,$$ for any shift-invariant space $S$ of $L^2({\mathbb{R}}^n)$.
\end{proof}

The next lemma follows the ideas from \cite{BG04}. It states that the angle between two shift-invariant spaces is the essential supremum of the angles between the fiber spaces.

\begin{lemma}\label{lem-ess-sup}
Let $U$ and $V$ be SISs of $L^2({\mathbb{R}}^n)$. Then,
$$\textbf{c}[U,V]=\textnormal{ess-sup }\{\textbf{c}[J_U(\omega),J_V(\omega)]\,:\,\omega\in\mathbb{T}^n\}.$$
\end{lemma}

\begin{proof}
Given $f\in V$, by Proposition \ref{proy}, we have for a.e. $\omega\in\mathbb{T}^n,$
$$\tau (P_U\big\vert_Vf)(\omega)=\tau (P_UP_Vf)(\omega)=P_{J_U(\omega)}P_{J_V(\omega)}(\tau f(\omega))=P_{J_U(\omega)}\big\vert_{J_V(\omega)}(\tau f(\omega)).$$
This shows that $P_{J_U(\omega)}\big\vert_{J_V(\omega)}$ is the range operator corresponding
to the shift-preserving operator $P_U\big\vert_{V}$ in the shift-invariant space $V$. What implies  that
\begin{equation}\label{norma-proy-fibras}
\|P_U\big\vert_V\|=\text{ess-sup }\{\|P_{J_{U}(\omega)}\big\vert_{J_{V}(\omega)}\|\,:\,\omega\in\mathbb{T}^n\}
\end{equation}
(see \cite{Bow00} for the definition and properties of shift-preserving operators).

Using (\ref{norma-proy-fibras}), Proposition \ref{cerrados1} and Lemma \ref{Js}, we obtain
\begin{align*}
\textbf{c}[U,V]&=\textbf{c}_0[U\ominus V,V\ominus U]=\|P_{U\ominus V}\big\vert_{V\ominus U}\|\\
&=\text{ess-sup }\{\|P_{J_{U\ominus V}(\omega)}\big\vert_{J_{V\ominus U}(\omega)}\|\,:\,\omega\in\mathbb{T}^n\}\\
&=\text{ess-sup }\{\|P_{J_{U}(\omega)\ominus J_{V}(\omega)}\big\vert_{J_{V}(\omega)\ominus J_{U}(\omega)}\|\,:\,\omega\in\mathbb{T}^n\}\\
&=\text{ess-sup }\{\textbf{c}_0[J_{U}(\omega)\ominus J_{V}(\omega),J_{V}(\omega)\ominus J_{U}(\omega)]\,:\,\omega\in\mathbb{T}^n\}\\
&=\text{ess-sup }\{\textbf{c}[J_{U}(\omega),J_{V}(\omega)]\,:\,\omega\in\mathbb{T}^n\}.
\end{align*}
\end{proof}

With the above results, we are able to prove the main theorem of this section.

\begin{proof}[\textit{Proof of Theorem \ref{teofinal}}]
By Lemma \ref{Js}, it follows that  $\tau\Phi(\omega)$ and
$\tau\Phi' (\omega)$ are frames for $J_{U}(\omega)\ominus J_{V}(\omega)$ and $J_{V}(\omega)\ominus J_{U}(\omega)$ respectively,
for a.e. $\omega\in\mathbb{T}^n$.

Thus, using Theorem \ref{teo-c}, we obtain
$$\mathbf{c}[J_{U}(\omega),J_{V}(\omega)]=\|(\mathcal{G}_{\Phi' }(\omega)^{\dagger})^{\frac{1}{2}}\mathcal{G}_{\Phi,\Phi' }(\omega)(\mathcal{G}_{\Phi}(\omega)^{\dagger})^{\frac{1}{2}}\|\quad\text{for a.e. }\omega\in\mathbb{T}^n.$$

Hence, by Lemma \ref{lem-ess-sup},
\begin{equation}\label{last-ec}
\textbf{c}[U,V]=\text{ess-sup }\{\|(\mathcal{G}_{\Phi' }(\omega)^{\dagger})^{\frac{1}{2}}\mathcal{G}_{\Phi,\Phi' }(\omega)(\mathcal{G}_{\Phi}(\omega)^{\dagger})^{\frac{1}{2}}\|\,:\, \omega\in\mathbb{T}^n\}.
\end{equation}

The proof of the theorem follows from (\ref{last-ec}) and Proposition \ref{cerrados1}.
\end{proof}

Next we provide an example of two shift-invariant spaces whose sum is not closed. In order to prove that,  we compute the Friedrichs angle  between the subspaces.

\begin{example}\label{ejclau}

Let $\varphi_1\in L^2({\mathbb{R}})$ be given by
$$\hat{\varphi_1}(\omega)=
\begin{cases}
\cos(2\pi\omega)   &\text{ if } 0\leq \omega <1\\
\sin(2\pi\omega) &\text{ if } 1\leq \omega <2\\
0 &\text{otherwise},
\end{cases}$$
and $\varphi_2, \varphi_3\in L^2({\mathbb{R}})$ satisfying that $\hat{\varphi_2}(\omega)=\chi_{[2,3)}(\omega)$ and $\hat{\varphi_3}(\omega)=\chi_{[3,4)}(\omega)$.
Define $U=S(\varphi_1, \varphi_2, \varphi_3)$.\\
Consider now $\varphi_0,\varphi_4\in L^2({\mathbb{R}})$, such that $\hat{\varphi_0}(\omega)=\chi_{[0,1)}(\omega)$ and $\hat{\varphi_4}(\omega)=\chi_{[\frac{5}{2}, \frac{7}{2})}(\omega)$, set $V=S(\varphi_0, \varphi_4).$

We will prove that $U+V$ is not closed using Theorem \ref{teofinal}.

Let $\{e_k\}_{k\in{\mathbb{Z}}}$ be the standard basis for $\ell^2({\mathbb{Z}})$. Then,
 $ \tau\varphi_1(\omega)=\cos(2\pi\omega)e_0+\sin(2\pi\omega)e_1$,  $ \tau\varphi_2(\omega)=e_2$,  $ \tau\varphi_3(\omega)=e_3$,
  $ \tau\varphi_0(\omega)=e_0$,  $ \tau\varphi_4(\omega)=e_3\chi_{[0,\frac{1}{2})}(\omega)+e_2\chi_{[\frac{1}{2},1)}(\omega)$. So, we have that for a.e. $\omega\in[0,1)$,
$$J_U(\omega)\ominus J_V(\omega)=\text{span}\{ \tau\varphi_1(\omega), \tau\varphi_5(\omega)\}\quad\text{and}\quad J_V(\omega)\ominus J_U(\omega)=\text{span}\{ \tau\varphi_0(\omega)\},$$ where $\hat{\varphi_5}(\omega)=\chi_{[2,\frac{5}{2})}(\omega)+\chi_{[\frac{7}{2},4)}(\omega)$.
Thus, by Lemma \ref{Js}, it follows that $U\ominus V=S(\varphi_1, \varphi_5)$ and $V\ominus U=S(\varphi_0)$.

Let $\Phi:=\{\varphi_1,\varphi_5\}$ and $\Phi':=\{\varphi_0\}$, then
$$
\mathcal{G}_{\Phi'}(\omega)=1\quad\mathcal{G}_{\Phi}(\omega)=
\begin{pmatrix}
1& 0\\
0 & 1
\end{pmatrix}\quad\text{and}\quad \mathcal{G}_{\Phi,\Phi' }(\omega)=(\cos(2\pi\omega),  0).
$$

Therefore
\begin{align*}
\textbf{c}[U,V]&=\textnormal{ess-sup } \{\|(\mathcal{G}_{\Phi'}(\omega)^{\dagger})^{\frac{1}{2}}\mathcal{G}_{\Phi,\Phi' }(\omega)(\mathcal{G}_{\Phi}(\omega)^{\dagger})^{\frac{1}{2}}\|\,:\, \omega\in[0,1)\}\\
&=\textnormal{ess-sup } \{|\cos(2\pi\omega)|\,:\,\omega\in[0,1)\}=1.
\end{align*}
Hence, by Theorem \ref{teofinal}, $U+V$ is not closed.

\end{example}

\vspace{13pt}
\centerline{ACKNOWLEDGEMENT}
\vspace{13pt}

\noindent We  thank J.L. Romero  for pointing out reference \cite{Deu95} that originated the content of Section \ref{section6}. We also thank the referee for his comments and suggestions that helped to improve the presentation.

redpajama_set_name: RedPajamaArXiv


1 Chronicles 7:20 - The descendants of Ephraim: Shuthelah, Bered his son, Tahath his son, Eleadah his son, Tahath his son,.
1 Chronicles 7:20 (GW) Ephraim's son was Shuthelah. Shuthelah's son was Bered. Bered's son was Tahath. Tahath's son was Eleadah. Eleadah's son was Tahath.
1 Chronicles 7:20 (NCV) These are the names of Ephraim's descendants. Ephraim's son was Shuthelah. Shuthelah's son was Bered. Bered's son was Tahath. Tahath's son was Eleadah. Eleadah's son was Tahath.
1 Chronicles 7:20 (NIRV) Here are the members of the family line of Ephraim. Shuthelah was Ephraim's son. Bered was Shuthelah's son. Tahath was Bered's son. Eleadah was Tahath's son. Tahath was Eleadah's son.
- Here is no account either of Zebulun or Dan. We can assign no reason why they only should be omitted; but it is the disgrace of the tribe of Dan, that idolatry began in that colony which fixed in Laish, and called it Dan, ( Judges 18 ) and there one of the golden calves was set up by Jeroboam. Dan is omitted, ( Revelation 7 ) . Men become abominable when they forsake the worship of the true God, for any creature object.
1 Chronicles 7:1-5 . SONS OF ISSACHAR.
1. Jashub--or Job ( Genesis 46:13 ).
2. whose number was in the days of David two and twenty thousand and six hundred--Although a census was taken in the reign of David by order of that monarch, it is not certain that the sacred historian had it in mind, since we find here the tribe of Benjamin enumerated [ 1 Chronicles 7:6-12 ], which was not taken in David's time; and there are other points of dissimilarity.
5. fourscore and seven thousand--exclusive of the 58,600 men which the Tola branch had produced ( 1 Chronicles 7:24 ), so that in the days of David the tribe would have contained a population of 45,600. This large increase was owing to the practice of polygamy, as well as the fruitfulness of the women. A plurality of wives, though tolerated among the Hebrews, was confined chiefly to the great and wealthy; but it seems to have been generally esteemed a privilege by the tribe of Issachar, "for they had many wives and sons" [ 1 Chronicles 7:4 ].
1 Chronicles 7:6-12 . OF BENJAMIN.
6. The sons of Benjamin--Ten are named in Genesis 46:21 , but only five later ( 1 Chronicles 8:1 , Numbers 26:38 ). Perhaps five of them were distinguished as chiefs of illustrious families, but two having fallen in the bloody wars waged against Benjamin ( Judges 20:46 ), there remained only three branches of this tribe, and these only are enumerated.
7. the sons of Bela--Each of them was chief or leader of the family to which he belonged. In an earlier period seven great families of Benjamin are mentioned ( Numbers 26:38 ), five of them being headed by these five sons of Benjamin, and two descended from Bela. Here five families of Bela are specified, whence we are led to conclude that time or the ravages of war had greatly changed the condition of Benjamin, or that the five families of Bela were subordinate to the other great divisions that sprang directly from the five sons of the patriarch.
12. Shuppim also, and Huppim--They are called Muppim and Huppim ( Genesis 46:21 ) and Hupham and Shupham ( Numbers 26:39 ). They were the children of Ir, or Iri ( 1 Chronicles 7:7 ).
of Aher--"Aher" signifies "another," and some eminent critics, taking "Aher" as a common noun, render the passage thus, "and Hushim, another son." Shuppim, Muppim, and Hushim are plural words, and therefore denote not individuals, but the heads of their respective families; and as they were not comprised in the above enumeration ( 1 Chronicles 7:7 1 Chronicles 7:9 ) they are inserted here in the form of an appendix. Some render the passage, "Hushim, the son of another," that is, tribe or family. The name occurs among the sons of Dan ( Genesis 46:23 ), and it is a presumption in favor of this being the true rendering, that after having recorded the genealogy of Naphtali ( 1 Chronicles 7:13 ) the sacred historian adds, "the sons of Bilhah, the handmaid, who was the mother of Dan and Naphtali." We naturally expect, therefore, that these two will be noticed together, but Dan is not mentioned at all, if not in this passage.
1 Chronicles 7:13 . OF NAPHTALI.
13. Shallum--or Shillem ( Genesis 46:24 ).
sons of Bilhah--As Dan and Naphtali were her sons, Hushim, as well as these enumerated in 1 Chronicles 7:13 , were her grandsons.
1 Chronicles 7:14-40 . OF MANASSEH.
14,15. The sons of Manasseh--or descendants; for Ashriel was a grandson, and Zelophehad was a generation farther removed in descent ( Numbers 26:33 ). The text, as it stands, is so confused and complicated that it is exceedingly difficult to trace the genealogical thread, and a great variety of conjectures have been made with a view to clear away the obscurity. The passage ( 1 Chronicles 7:14 1 Chronicles 7:15 ) should probably be rendered thus: "The sons of Manasseh were Ashriel, whom his Syrian concubine bare to him, and Machir, the father of Gilead (whom his wife bare to him). Machir took for a wife Maachah, sister to Huppim and Shuppim."
21. whom the men of Gath . . . slew, &c.--This interesting little episode gives us a glimpse of the state of Hebrew society in Egypt; for the occurrence narrated seems to have taken place before the Israelites left that country. The patriarch Ephraim was then alive, though he must have arrived at a very advanced age; and the Hebrew people, at all events those of them who were his descendants, still retained their pastoral character. It was in perfect consistency with the ideas and habits of Oriental shepherds that they should have made a raid on the neighboring tribe of the Philistines for the purpose of plundering their flocks. For nothing is more common among them than hostile incursions on the inhabitants of towns, or on other nomad tribes with whom they have no league of amity. But a different view of the incident is brought out, if, instead of "because," we render the Hebrew particle "when" they came down to take their cattle, for the tenor of the context leads rather to the conclusion that "the men of Gath" were the aggressors, who, making a sudden foray on the Ephraimite flocks, killed the shepherds including several of the sons of Ephraim. The calamity spread a deep gloom around the tent of their aged father, and was the occasion of his receiving visits of condolence from his distant relatives, according to the custom of the East, which is remarkably exemplified in the history of Job ( Job 2:11 ; compare John 11:19 ).


Choose the right used GMC that fits your bill! Search for a used GMC in Los Angeles, California by price, distance and make.
Los Angeles, CA auto insurance quotes. Save money when you compare auto insurance quotes.
Compare new car price quotes from multiple auto dealers near Los Angeles, CA.
Find used automotives for sale in Los Angeles, CA.


Congratulations Nancy Staves! Nancy, a volunteer in our Plattsburgh office, was honored for her time and dedication. She was recognized for more than ten years of service and the most administrative hours. We thank Nancy for sharing her time and talents with us.


Junichi Sato's beautiful world of Aqua is beckoning you to come visit in ARIA The ANIMATION!
In the early 24th century, Mars has been terraformed by mankind into a sparkling planet covered in water. Akari Mizunashi, at the age of 15, has left everything behind to travel to that reborn world, now known as Aqua. More than anything, Akari wants to be an "undine" – a female gondolier who navigates the canals of the Aquan city of Neo-Venezia. As she begins her training with the prestigious Aria Company, will she be up to the challenges that await her on the path to achieve her dream?
Contains all 13 episodes of season 1 of ARIA and packed full of amazing extras!
Special Features: Apprentice Undines Interview Parts 1-2, Primas Interview Parts 1-2, SatoJun's "Venice. I'm Sorry!" Parts 1-4, 30-second & 15-second Commercials, US Season 1 Trailer, Clean Endings 1-2, Season 2 Preview, Trailers.
© 2005 Kozue Amano / Mag Garden - ARIA Company.


Well, apparently this one runs for four weeks, so feel free to knock yourself out.
I have a seeecret project for the Expo. Is the site ready for people to start setting up booths?
Not sure what you're saying, but I assume you mean it's started. I've found the site and set up a login, but I'm not sure exactly when the expo ends. I've only now started on my project and it'll take a good week or so to even be presentable.
Well, this year's QExpo is going to last a month, it seems. It started on the 15th of August (if I remember well), so it should go on till mid-September.


Broadway is a small road that runs across the Wentlooge Level.
The road Carys lives on.
Imperial Park Bypass (Sleeper, Meat, A Day in the Death). Road (2.8 miles, 8m) : Head north on the Broadway over the railway into Marshfield. Turn right onto Ty Mawr Lane and then at the end turn left onto Church Lane. This then leads out to the roundabout at the top of the bypass. It is possible to park alongside the bypass on the side of the roundabout.
St. Mary's Church (The End of Time): 0.83 miles.
Office Building (Prisoner of the Judoon): 1.18 miles.
Heron Marsh (Sarah Jane Adventures Series 3): 1.59 miles.
Vaendre Hall (Doctor Who Christmas Special 2014 Filming): 1.59 miles.
A48 (Castleton) (The Eleventh Hour): 1.60 miles.
Random Shoes : Eugene's death.


This vellum snowman features the Stampin' Up! set Snow Much Fun, a set that truly is SNOW much fun! I find myself reaching for it time after time, and still can't get enough of the two happy snowmen wearing their winter boots, accompanied by the cheery winter birds. These characters never fail to make me smile.
The tutorial to make this card can be purchased in a PDF format through my Etsy store, here: http://www.etsy.com/shop/paperseedlings The techniques used to create the card are fairly simple and result in a nice icy effect.
I've never tried vellum. I love the snowman too.
Vellum is so much fun -- and so elegant to play with. You should really give it a try, Christina!
Oh he's adorable! The blues make me shiver!! Great job!
I hope you're warmed up by now. :>) Thanks for the nice compliment!
I love the card so cute!
this is such a cute snowman!


text:
It's Time to Win: The Orient Limited Edition Cal. 469 40th Anniversary Watch Giveaway
The 40th Anniversary of the 469 Caliber was definitely a milestone in Orient's rich history. Did you know that over 100 million 469 caliber movements have been produced to date? It's Orient's longest produced movement and for its 40th Anniversary years ago they created two models to celebrate it. They're reminiscent of the popular Orient King Diver that was extremely popular in the 1970s. These were limited edition: only 3000 total were manufactured, and each were individually serial numbered: 1000 of the black IP version and 2000 in brushed stainless steel.
We're giving away the EM7F001B, in IP black, number 0369/1000
How do you enter? It's simple. Sign up for our Newsletter! (The easiest way is through the sign up form on the bottom right of our home page www.orientwatchusa.com). Our newsletter is the source for all the new releases and we give special deals to our subscribers (early access sales, free stuff, etc.).
Be sure to sign up with your email by July 31, 2013, though! Look below for the official rules!
ORIENT WATCH USA 469 ANNIVERSARY WATCH GIVEAWAY/CONTEST RULES AND REGULATIONS
1. Eligibility: The Orient Watch USA 469 Anniversary Watch Giveaway (the "Promotion") begins at 1:00:00 PM Pacific Standard Time (PST) on July 11, 2013 and ends at 11:59:59 PM PST on July 31th, 2013 (the "Promotion Period"). The Promotion is only open to legal residents of the United States (50 states and District of Columbia, but specifically excluding the territory of Puerto Rico) who are eighteen (18) years old and over as of the date of entry. Employees of Orient Watch USA ("Sponsor"), and each of their, affiliates, agents, advisors, advertising and promotion agencies, retail jewelry/watch accounts, official judges and members of the immediate families and households of each (collectively "Sponsor Entities") are not eligible to enter, participate or win. This Promotion is governed by U.S. law and is subject to all applicable federal, state and local laws and regulations. Void where prohibited or restricted by law. By participating in or entering this Promotion, each entrant unconditionally agrees to be bound by and accepts these official rules (the "Official Rules") and the decisions of the Sponsor as binding and final in all respects.
2. How to Enter: To enter the Promotion, sign up for Orient Watch USA's email newsletter (can be done via www.orientwatchusa.com/newsletter). All entries (or email newsletter sign ups) must be done by no later than 11:59:59 PM PST on July 31th, 2013, and must be complete in their entirety to be eligible for entry into the Promotion. Entries can only be submitted online via email newsletter sign up. Limit one (1) entry per person during the Promotion Period. Entries by third parties, commercial or automatic sweepstakes entry services, scripts, bots, macros, or users in possession of multiple Facebook identities are prohibited. Any entry received by the aforementioned methods will be deemed ineligible. Sponsor assumes no responsibility for lost, late, damaged, misdirected, illegible, ineligible, or incomplete entries.
3. Winner Selection: One (1) eligible entrant who participates in the Orient Watch USA 469 Anniversary Watch Giveaway will be selected at random. The chance of winning depends on the number of entrants. The potential winner will be notified by email within one week of the selection. The potential winner will be required to sign and return an Affidavit of Eligibility and Release of Liability/Publicity within 3 business days of initial notification or the potential winner will forfeit the prize and a new winner will be chosen at random from among the remaining eligible entries. Acceptance of a prize constitutes irrevocable consent to the use of his/her voice, name, likeness, biographical information and/or photograph by Sponsor for promotional or advertising purposes in connection with the Promotion, in any and all media worldwide, including, without limitation, in print, TV, radio, cable, satellite and online promotions, without limitation, restriction, condition, notice, review, approval or additional compensation, except where and to the extent prohibited by law.
4. Prize: A total of one (1) prize will be awarded, consisting of one (1) Orient Watch from the Orient Watch USA 469 Anniversary collection, selected at the discretion of Sponsor. Approximate retail value ("ARV") of the prize: $250-$495 USD. The prize will be delivered free of charge. No redemption, substitution or transfer of the prize permitted. Each winner is responsible for all taxes associated with the acceptance and use of the prize.
5. Limitation of Liability: By entering this Promotion, participants agree to release Sponsor and Sponsor Entities from any liability arising from any injury, death, loss or damage of any kind resulting from or in connection with entrance and participation in this Promotion, or the awarding, acceptance, possession, use or misuse of the prize or portion thereof awarded in connection with this Promotion. By entering, you also agree that Sponsor and Sponsor Entities are not responsible for technical, computer, mechanical, printing, typographical, human or other errors relating to or which may occur in connection with the Promotion, including, without limitation errors which may occur in the administration of the Promotion, the processing of entries, the announcement of the prizes or in any Promotion related materials? or for transactions or entries that are incomplete, inaccurate, garbled, fail to enter into the processing system, or are processed, reported, or transmitted late or incorrectly or are lost for any reason including computer, telephone, human or other error? or for electronic, computer, or telephonic malfunction or error, including inability to access any web site associated with this promotion. If in the Sponsor's opinion, there is any suspected or actual evidence of electronic or nonelectronic tampering with any portion of the Promotion, or if a computer virus, bug, unauthorized intervention, fraud, or technical difficulties or failures compromise or corrupt or affect the administration, integrity, security, fairness, or proper conduct of the Promotion, Sponsor reserves the right in its sole discretion to disqualify any individual who tampers with the entry process, void any entries submitted fraudulently, modify or suspend the Promotion, or terminate the entire Promotion or that portion of the Promotion relating to the suspected or actual tampering of the Promotion and conduct a random drawing to award the prizes using all nonsuspect, eligible entries received as of the date of such action. Any attempts by an individual to access any web site associated with this Promotion via robotic, automatic entry devices, programs such as, but not limited to, script, macro or any other automated means, or other unauthorized entry will void all such entries by such methods. Any attempt by an entrant or any other individual to deliberately damage any web site or undermine the legitimate operation of the Promotion is a violation of criminal and civil laws and should such an attempt be made, the Sponsor reserves the right to seek damages and other remedies from any such person to the fullest extent permitted by law. In addition, Sponsor nor Sponsor Entities have any liability or responsibility for or related to the prize and there are no warranties, representations or guarantees, express or implied, in fact or in law, relative to any portion of the prize other than the warranty provided by Orient Watch USA. Prize does not include any and all other expenses not specified herein. By registering, entrants hereby release Facebook from any liability. This promotion is in no way sponsored, endorsed or administered by, or associated with Facebook.
6. Rules and Winners: For a copy of the Official Rules or the names of the winner, send a self addressed, stamped envelope to Orient Watch USA, 631 S. Olive Street #422, Los Angeles California 90014 by July 11, 2013. (VT residents may omit postage from rule request).
7. This Promotion is sponsored by Orient Watch USA, located at 631 S. Olive Street #422, Los Angeles California 90014.
Back to Orient
Other stories from the blog
It's Time to Win: The Orient Limited Edition Cal. 469 40th Anniversary | Orient Watch USA

redpajama_set_name: RedPajamaCommonCrawl


It's a rare parent who doesn't have to deal with a child's temper tantrum, meltdown, or other challenging behavior at some point. Why do young children act up, and what triggers the behavior? This article will help to answer those questions and assist you in developing a plan that can improve your child's behavior.
The first thing to realize is that all behavior has a purpose: communication. Children who don't have an appropriate way to express ideas and feelings may cry, hit, or pout to let you know they need something, such as food; want something, such as a favorite toy; or want to avoid something, such as going to bed. Children may act up for other reasons as well. They may have limited social skills, may have learned that misbehaving is more effective than language, may be physically uncomfortable, may not know what's expected of them, or may be faced with unreasonable expectations.
The situations may be different, but one theme unites them: Children engage in challenging behavior because it works for them.
The first step in helping children improve their behavior is to determine what they are trying to communicate and when the behavior is most likely to occur. Keeping a record of when and where the behavior happens can help you identify patterns. Be sure to note what was happening just prior to the behavior, what you observed, and what happened afterward. Is there a certain time of day or a particular activity that seems to be more difficult for your child?
Once you have determined the purpose of the behavior, you can create a plan to change it. This behavior plan should have three parts: prevention, education, and consistency.
Try these strategies to help increase desired behaviors and decrease unwanted ones.
Express clear, realistic expectations to your child. For example, if your daughter is not able to pick up all her toys by herself but could help you with the task, you might express that realistic expectation by saying, "I'll pick up 10 toys and you'll pick up 10 toys." After the toys are picked up, praise your child for her cooperation.
Set a few clear household rules. Be sure to state the rules in a positive manner, explaining what you want your child to do, not what you don't want. For instance, you might say, "use gentle touch" rather than "no hitting." Review the rules daily and make sure your child understands what is expected. If you also have consequences for breaking the rules, make them clear and be sure to give them consistently.
Teach your child new skills for communicating and participating in routines or expectations.
Teach your child the language to use to in order to meet his or her needs. For example, if your son tries to grab a toy from another child, teach him how to obtain that toy in an appropriate manner. You might say, "Zach, please ask Jenny to give you the blocks when she is finished playing with them. Now say thank you." If your daughter whines when she wants a drink of juice, you could say, "Sally, say 'juice please.'" (For a nonverbal child, teach the sign or picture to use.) Once you have given her the words and she has the ability to use them, tell her you will expect her to use her words to ask for juice next time and ignore the whining. Give Sally the juice only when she uses her words or signs.
Children also need to learn and practice problem solving skills. If your son throws his plate to indicate that he wants to be done with a meal, teach him the words or sign language for "all done." Give him several opportunities to practice, and then be sure that he can no longer escape mealtime by throwing his plate. Remember to praise your child when he uses his words or signs to let you know he's finished with a meal.
Be consistent in making sure your child's challenging behaviors are no longer effective.
Respond to your child's unacceptable actions with consequences that are practical and logically related to the behavior. For example, if your daughter is using a toy in an inappropriate way, calmly remind her how it is to be used and make it clear that if she persists, the toy will be put away. If she continues using the toy inappropriately, calmly remove the toy. Your child will learn that you expect her to play with toys in an appropriate manner. Logical consequences teach children that they have control over their own behavior.
By understanding why children have challenging behaviors, you can plan effective responses. Prevention strategies, new skills, and proper reinforcement of desired behaviors can all help. As you develop your behavior plan, make sure that it will fit with your family and that you are committed to the work and energy it will take to follow through.
To provide your child with consistency, you also may find it valuable to involve childcare providers, preschool teachers, extended family members, or other people who care for your child in a plan to address challenging behaviors. If your child has an Individualized Family Service Plan (IFSP) or Individualized Education Program (IEP), you can ask the team members how to create and follow through on a behavior plan.
Change won't happen overnight, but as children learn that challenging behaviors no longer work for them, they will begin to use the new skills you have taught them. As you lay the foundation for helping them learn new ways to interact and behave, the quality of their relationships will improve and your time with your child will become less stressful and more enjoyable.
Won't or Can't: Why Don't Children Behave?
There are two ways to look at challenging behavior. One is to conclude that the child won't behave. In that case, the first response may be to punish. Punishment, however, seldom teaches a new skill or behavior. The other view is that the child can't behave. In that case, the response is to teach the child the skills that would help meet his or her needs in a more appropriate manner.


text:
Argelia Garcia (Mexico; 1978) es una actriz mexicana de televisión que actualmente trabaja para Telemundo en los Estados Unidos, participando en varias telenovelas de dicha empresa. Es la mama del también actor de telenovelas infantiles Jorge Eduardo García.

Filmografía

Televisión

 La suerte de Loli (2021)
 100 días para enamorarnos (2020) --- Ana Hussman
 Al otro lado del muro (2018) ---Margarita
 Mariposa de barrio (2017) --- Luisa Rivera
 Eva, la trailera (2016) ---- Celeste Mejia
 Dama y obrero (2013) --- Elena
 Amar de nuevo (2011) --- Leticia
 Capadocia (2010) Mama avión

Referencias

Enlaces externos

redpajama_set_name: RedPajamaWikipedia


text:
Bruce Backhouse
Posted on 2007/03/06 (2012/10/25) by Jacqueline Nurse
OPENING: Saturday 17 March 2007 @ 12.30
Exhibition concludes 21 April
Bruce Backhouse began his art career many years ago at Rhodes University where he studied Fine Art for two years before deciding that the world of advertising suited him better. He remained in advertising for the next twenty-five years until, four years ago, he returned to his earliest aspiration to pursue a career as a painter.
Although Backhouse's style is loose and painterly, he favours a cartoon-like rendition of figures and objects, often with a strong emphasis on the drawn outline. In 2006 Backhouse decided to try his hand at printmaking as a way of achieving a finer, more lyrical outline and he joined printmaker Niall Bingham's Sunday courses at the David Krut Print Workshop (DKW). Backhouse soon found a niche in etching and aquatint techniques, a combination that allowed him to concentrate on his characteristic quick-drawn line, whilst retaining his atmospheric backgrounds. Backhouse was subsequently invited to continue his collaboration with Bingham at DKW and, in addition to the techniques he had already acquired, now started to experiment with monotypes, through which he reintroduced colour into his oeuvre. Backhouse's monotypes have retained the charm of his earlier watercolour paintings but have propelled him into new areas of visual exploration. The intensity of colour achievable with monotypes has been a source of excitement for the artist, as has the ability to combine more than one technique in a print. Fine line can be achieved through etching and dry point, whilst moody, textured greys are the product of spitbite and aquatint. Colours are brighter and can be blended in a way that is simply not possible with watercolours.
Backhouse has continued to explore the genre of landscape in his new body of work but, although his subject matter comes from the rocky outcrops and koppies that dot the vast Karoo landscape, he paints entirely from memory, eschewing geographical accuracy. Backhouse describes his artistic mission as a desire to change the pervasive perception of landscape painting since he believes that this genre has been rehashed to the extent that it can no longer engage artist or viewer on any meaningful level. He prefers to use the landscape as a stage for dynamic social interaction rather than as static topographical record.
In spite of the aforementioned explanations, Backhouse likes to keep his art as uncomplicated as possible. One has the impression, when speaking to the artist, that he feels genuinely privileged to have been granted the opportunity to immerse himself in his art and that his over-riding intention is to uplift his audience rather than deliver overly profound messages.
Julian Opie, New Editions